Question
Write a value of $\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$=\frac{\text{e}^{-\text{x}}}{\text{e}^{-\text{x}}+1}\text{dx}$
Let $\text{e}^{-\text{x}}+1=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=-\log|1+\text{e}^{-\text{x}}|+\text{C}$

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